Lesson 5

It is easy to verify that events \(A_{12}, A_{23}, A_{13}\) are pairwise independent: \[\Pr[A_{12}\cap A_{23}] = \frac{1}{4}=\Pr[A_{12}]\cdot\Pr[A_{23}] \ .\] However, \[\Pr[A_{12}\cap A_{23}\cap A_{13}] = \frac{1}{4}\neq \frac{1}{8}=\Pr[A_{12}]\cdot\Pr[A_{23}]\cdot\Pr[A_{13}] \ .\] The lack of mutual independence (3-independence) can be seen almost without calculations: \[\Pr[A_{13}\mid A_{12}\cap A_{23}]=1 \neq \Pr[A_{13}] \ .\]